Radial and Transverse Forces Beer and Johnston 9th ed. 12.70

[jaredogden]

Homework Statement

The 3 kg collar B slides on the frictionless arm AA'. The arm is attached to drum D and rotates about O in a horizontal plane at the rate dθ/dt = 0.75t, where dθ/dt and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of 0.5 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA'.

Homework Equations

F_r = m(d²θ/dt² - r(dθ/dt)²)
F_θ = m(rd²θ/dt² + 2(dr/dt)(dθ/dt))
ƩF = ma

The Attempt at a Solution

F_r = -T
r = (0.5 m/s)t
dr/dt = 0.5 m/s
d²r/dt = 0 m/s²
dθ/dt = (0.75 m/s)t
d²θ/dt = 0.75 m/s²

F_r = 3 kg(0 - 0.5t(0.75t)²)
F_r = -0.844t³

F_θ = 3 kg(0.5t(.75) + 2(0.5)(0.75t))
F_θ = 3.375t

Now if F_r = -T then I can substitute -T for F_r. Therefore
T = 0.844t³

I'm not sure where to go after this one.. I need to somehow solve for t I know that but I'm just not sure because I have two equations and three unknowns.

Thanks ahead of time for any help.

 

[KataKoniK]

Hello,

Thank you for posting your problem on the forum. I would like to offer some suggestions for solving this problem.

Firstly, it is important to carefully read and understand the problem statement. From the given information, we know that the collar B is sliding on the frictionless arm AA' and that the arm-drum assembly is rotating at a constant rate. The collar is also moving outward from O with a constant speed of 0.5 m/s. We are asked to determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by the arm AA'.

Based on the given information, we can write the following equations:

ƩF = ma (Newton's second law)
Fr = -T (equilibrium condition for the collar B)
Fθ = maθ (Newton's second law in the tangential direction)

We can also use the given information to determine the values of the variables at t = 0:

r = 0 (initial position of the collar)
dr/dt = 0.5 m/s (constant speed of the collar)
dθ/dt = 0.75 rad/s (constant angular velocity)

To solve for the time t, we need to find a relationship between the tension T and the horizontal force exerted on the collar by the arm AA'. We can do this by equating the equations for Fr and Fθ:

Fr = Fθ
- T = maθ
- T = 3 kg (r(dθ/dt)2 - d2r/dt2)
- T = 3 kg (0.5t(0.75t)2 - 0)
- T = 0.844t3

Now, we can use this equation to solve for t when T = Fθ:

0.844t3 = 3.375t
t = (3.375/0.844)1/2
t = 1.69 s

Therefore, the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on the collar by the arm AA' is 1.69 seconds.

I hope this helps. Let me know if you have any further questions.